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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Spring Pendulum

  • question_answer
    A mass M is suspended by two springs of force constants K­1 and K2 respectively as shown in the diagram. The total elongation (stretch) of the two springs is   [MP PMT 2000; RPET 2001]

    A)            \[\frac{Mg}{{{K}_{1}}+{{K}_{2}}}\]

    B)            \[\frac{Mg\,({{K}_{1}}+{{K}_{2}})}{{{K}_{1}}{{K}_{2}}}\]

    C)            \[\frac{Mg\,{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\]

    D)            \[\frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}Mg}\]

    Correct Answer: B

    Solution :

                       For series combination \[{{k}_{eq}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]            \[F={{k}_{eq}}x\Rightarrow mg=\left( \frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} \right)x\]\[\Rightarrow x=\frac{mg({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}\]


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